Question

What is the pH of the resulting solution if 30. mL of 0.432 M methylamine, CH3NH2, is added to 15 mL of 0.234 M HCl? Assume that the volumes of the solutions are additive. Ka = 2.70 × 10-11 for CH3NH3+ ?

Answer 1

HCl is an acid and it will react with the base. In these cases, the HCl is the limiting reagent and, at the end of the reaction, we have a solution with some weak base (the methylamine) and the salt of a weak base (the methylammonium chloride, CH3NH3Cl).

We will use the Henderson-Hasselbalch equation at the end to get the pH.

HCl ---> (0.234 mol/L) (0.015 L) = 0.00351 mol

CH3NH2 ---> (0.432 mol/L) (0.030 L) = 0.01296 mol

100% of the HCl is used up. what remains is this:

methylamine ---> 0.01296 mol - 0.00351 mol = 0.00945 mol

methylammonium ion ---> 0.00351 mol

we know,

pH = pKa + log (base/acid)

The pKa of CH3NH3+ ---> pKa = -log 2.70 × 10^-11 = 10.5686

pH = 10.5686 + log (0.00945/0.00351) = 10.5686 + 0.430

pH= 10.998